# Receiving node's clock=3+sending node clock, confused in further calculations

Background information-:

Here is the definitions about what the timestamps mean-: Value of original timestamp: 46

Value of transmit timestamp: 60

Time the packet arrived: 67

Thus Sending time = receive timestamp=original timestamp=59 - 46 = 13 milliseconds

Receiving time =time message arrived at sender-transmit timestamp= 67 - 60 = 7 milliseconds

Round-trip time =sending time+receiving time=13 + 7 = 20 milliseconds

Now say we Consider that receiving node's clock=3+sending node's clock.(I know it is really a basic level math but I am confused here as my initituion isn't working)

The the slide writes-:

Sending time = (56+3) – 46 = 10 + 3 Receiving time = 67 – (57+3) = 10 – 3 RTT = (10 + 3) + (10-3) = 20

Here is where I get confused. I personally believe that it should be-: sending time=(59+3)-46=13+3(as receiving time=3+sending time) receiving time=67+3-60=10

But I am wrong. But I can't comprehend what is written in that slide.

BTW here is the slide-: http://osnet.cs.nchu.edu.tw/powpoint/Computer%20Network93_1/Chapter%209.pdf

It is page number 67/86

I know this might be really a dumb question to ask, but I have been banging my head for hours and unable to find a solution of this. Please guide.

You are misinterpreting what they are saying (because they are not clear in how they say it). Their intent is that the clock at the receiver is $$3$$ ms ahead of the clock at the sender. So

• Receiver timestamp of $$59$$ occurs at the same time as $$56$$ on the Sender clock: $$59 = 56 + 3$$.
• Receiver timestamp of $$60$$ occurs that the same time as $$57$$ on the Sender clock: $$60 = 57 + 3$$.

So if the Receiver clock were synchonized to the Sender clock, the timestamps on its end would be $$56$$ and $$57$$, and the calculation would look like

• sending time $$= 56 - 46 = 10$$ ms.
• receiving time $$= 67 - 57 = 10$$ ms.
• RTT $$= 10 + 10 = 20$$ ms.

While the actual calculation with unsynchronized clocks is

• sending time $$= 59 - 46 = 13$$ ms.
• receiving time $$= 67 - 60 = 7$$ ms.
• RTT $$= 13 + 7 = 20$$ ms.

The error in the clock synchronization changes the sending time and receiving time in opposite directions. But when you add them to get the round-trip time, the changes cancel each other, and the round-trip time is not changed. The point is the round-trip time is unaffected by this almost-unavoidable error, while the two one-way times are inaccurate.